Term Rewriting System R:
[x, y]
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(c(s(x), y)) -> F(c(x, s(y)))
F(c(s(x), s(y))) -> G(c(x, y))
G(c(x, s(y))) -> G(c(s(x), y))
G(c(s(x), s(y))) -> F(c(x, y))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

G(c(s(x), s(y))) -> F(c(x, y))
G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), s(y))) -> G(c(x, y))
F(c(s(x), y)) -> F(c(x, s(y)))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





The following dependency pairs can be strictly oriented:

G(c(s(x), s(y))) -> F(c(x, y))
F(c(s(x), s(y))) -> G(c(x, y))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1, x2))=  x1 + x2  
  POL(G(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), y)) -> F(c(x, s(y)))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





Using the Dependency Graph the DP problem was split into 2 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Instantiation Transformation


Dependency Pair:

G(c(x, s(y))) -> G(c(s(x), y))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

G(c(x, s(y))) -> G(c(s(x), y))
one new Dependency Pair is created:

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Instantiation Transformation


Dependency Pair:

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

G(c(s(x''), s(y''))) -> G(c(s(s(x'')), y''))
one new Dependency Pair is created:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pair:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





The following dependency pair can be strictly oriented:

G(c(s(s(x'''')), s(y''''))) -> G(c(s(s(s(x''''))), y''''))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1, x2))=  x2  
  POL(G(x1))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 9
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Instantiation Transformation


Dependency Pair:

F(c(s(x), y)) -> F(c(x, s(y)))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(c(s(x), y)) -> F(c(x, s(y)))
one new Dependency Pair is created:

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Instantiation Transformation


Dependency Pair:

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(c(s(x''), s(y''))) -> F(c(x'', s(s(y''))))
one new Dependency Pair is created:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 8
                 ↳Polynomial Ordering


Dependency Pair:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))





The following dependency pair can be strictly oriented:

F(c(s(x''''), s(s(y'''')))) -> F(c(x'''', s(s(s(y'''')))))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  1 + x1  

resulting in one new DP problem.


Termination of R successfully shown.
Duration:
0:00 minutes